Question

please answer ASAP.. pleaseeeeeee​

Answer 1

Answer:

0

Step-by-step explanation:

Assuming it to be a real value problem, (one might assume x to be complex, solution might differ) break it into appropriate intervals.

1) When x < -8  

For this interval absolute (abs) value function can be expanded as follows.

-(x+3) - (4-x) = -(8+x)

solving it we get, x= -1; But x should be less than -8. Thus, it's not a solution.

2) When -8 < x < -3 , equation is

-(x+3) - (4-x) = (8+x)

solving it we get, x= -15; But x should be between -8 and -3. Thus, it's also not a solution.

3) When -3 < x < 4 , equation is

(x+3) - (4-x) = (8+x)

solving it we get, x= 9; But x should be between -3 and 4. Thus, it's also not a solution.

4) When  x > 4 , equation is

(x+3) - (-(4-x)) = (8+x)

solving it we get, x= -1; But x should be greater than 4. Thus, it's also not a solution.

Therefore, no value of x satisfies given equation.  

Hence, zero number of solution